$$\frac{125}{343}+\frac{5}{1}$$
$\frac{1840}{343}\approx 5.364431487$
$$\frac{125}{343}+5$$
$$\frac{125}{343}+\frac{1715}{343}$$
$$\frac{125+1715}{343}$$
$$\frac{1840}{343}$$
Show Solution
Hide Solution
$\frac{2 ^ {4} \cdot 5 \cdot 23}{7 ^ {3}} = 5\frac{125}{343} = 5.364431486880466$
